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CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANT

CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANT. A solution is saturated if it has a solid at equilibrium with its solute. The concentration of solute in a saturated solution is the largest that is normally possible. For a saturated solution of ionic solid in water:

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CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANT

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  1. CH 104: DETERMINATION OF A SOLUBILITY PRODUCT CONSTANT • A solution is saturated if it has a solid at equilibrium with its solute. The concentration of solute in a saturated solution is the largest that is normally possible. • For a saturated solution of ionic solid in water: • CxAy(s) = xCn+(aq) + yAm–(aq) • Solubility Product Constant = Ksp = [Cn+]x[Am–]y • Where • CxAy(s) is a slightly soluble ionic solid. • [Cn+] and [Am–] are the equilibrium concentrations of ions in moles per liter. • x and y are the stoichiometric coefficients from the balance reaction. • The solubility product constant (Ksp) is the equilibrium constant for the dissolution of a slightly soluble ionic compound at a specified temperature. • By convention the “concentration” of the solid, CxAy, is NOT used to calculate Ksp. (That is, the activity of a pure solid is 1.)

  2. SOLUBILITY PRODUCT CONSTANT • What is the reaction for a saturated solution of Bi2S3(s) in water? • Bi2S3(s) = 2Bi3+(aq) + 3S2–(aq) • What is the Ksp for this reaction? • Ksp = [Bi3+]2[S2–]3 • Notice the “concentration” of Bi2S3(s) is NOT used to calculate Ksp. • What is the reaction for a saturated solution of Ag2CrO4(s) in water? • Ag2CrO4(s) = 2Ag+(aq) + CrO42–(aq) • What is the Ksp for this reaction? • Ksp = [Ag+]2[CrO42–]

  3. SOLUBILITY PRODUCT CONSTANT • Calcium fluoride (CaF2) is slightly soluble in water. In a saturated solution the CaF2(s) is dissolving at the same rate that Ca2+(aq) and F–(aq) crystallize. That is, the solid and solute are at equilibrium. • CaF2(s) = Ca2+(aq) + 2F–(aq)

  4. CALCULATING SOLUBILITY FROM Ksp • A saturated solution is made by adding excess CaF2(s) to distilled water.What is the solubility of this CaF2(s) at 25° C? • Step #1: Write the balanced reaction and Ksp equation. • CaF2(s) = Ca2+(aq) + 2F–(aq) • Ksp = [Ca2+][F–]2 = 2.7x10–11 at 25° C • Step #2: The initial concentrations of Ca2+(aq) and F–(aq) are 0. The equilibrium concentrations of Ca2+(aq) and F–(aq) are given algebraic variables based on the stoichiometric coefficients from the balance reaction. Write these equilibrium concentrations of Ca2+(aq) and F–(aq). • [Ca2+] = x • [F–] = 2x

  5. CALCULATING SOLUBILITY FROM Ksp • Step #3: Use the Ksp equation to solve for [Ca2+] and [F–]. • Ksp = 2.7x10–11 = [Ca2+][F–]2 = (x)(2x)2 = 4x3 • x3 = 2.7x10–11 / 4 = 6.75x10–12 • [F–] = 2x = 3.8x10–4 M • Step #4: Solve for the solubility of CaF2(s). • One mole of Ca2+(aq) is produced for every mole of CaF2(s) that dissolves; therefore, • the solubility of CaF2(s) = [Ca2+] = 1.9x10–4 M.

  6. THE COMMON ION EFFECT • In the previous example the pure solid (CaF2(s)) was the only source of its dissolved ions (Ca2+(aq) and F–(aq)). • However, if the common ion F–(aq) is added it will react with Ca2+(aq) to decrease the solubility of CaF2(s). The new concentration of Ca2+(aq) is less than in the original equilibrium. And the new concentration of F–(aq) is greater than in the original equilibrium. This is called Le Châtelier’s principle. • CaF2(s) = Ca2+(aq) + 2F–(aq) • Ksp = [Ca2+][F–]2 = 2.7x10–11 at 25° C • Similarly, if the common ion Ca2+(aq) is added it will react with F–(aq) to the solubility ofCaF2(s). The new concentration of F–(aq) is than in the original equilibrium. And the new concentration of Ca2+(aq) is than in the original equilibrium. decrease less greater

  7. THE COMMON ION EFFECT • The common ion F–(aq) is added to a saturated solution of CaF2(s) in distilled water.What is the concentration of Ca2+(aq) in equilibrium with 1.0 M F–(aq) and CaF2(s) at 25° C? • Ksp = 2.7x10–11 = [Ca2+][F–]2 = [Ca2+](12) • [Ca2+] = 2.7x10–11 M • Compared to the previous example, did the concentration of Ca2+(aq) increase or decrease? • It decreased from 1.9x10–4 M to 2.7x10–11 M. • Did the concentration of F–(aq) increase or decrease? • It increased from 3.8x10–4 M to 1.0 M. • Does this agree with the common ion effect? • Yes. The concentration of Ca2+(aq) decreased. The concentration of F–(aq) increased. And the solubility of CaF2(s) decreased.

  8. THE SALT EFFECT • Common ions decrease the solubility of ionic solids. • In contrast, the presence of “uncommon” ions tends to increase solubility of ionic solids. This is called the “salt effect”, the “uncommon ion effect”, or the “diverse ion effect”. • Soluble uncommon ions increase the interionic attractions of a solution. As a result, these uncommon ions decrease the effective concentrations (or activities) of other solutes and increase the solubility of ionic solids.

  9. THE SALT EFFECT • The salt effect is not as striking as the common ion effect. • The presence of the common ion CrO42–(aq), from K2CrO4, decreases the solubility of Ag2CrO4 by a factor of 35. • In contrast, the presence of the uncommon ions K+(aq) and NO3–(aq), from KNO3, increase the solubility of Ag2CrO4 by a factor of only 0.25.

  10. CALCULATING Ksp FROM SOLUBILITY • In today’s experiment you will measure the solubility of potassium hydrogen tartrate (KOOC(CHOH)2COOH). • KOOC(CHOH)2COOH(s) = K+(aq) + –OOC(CHOH)2COOH(aq) • What is the Ksp for this reaction? • Ksp = [K+][–OOC(CHOH)2COOH] • You will make a saturated solution of KOOC(CHOH)2COOH in 0.10 M NaCl and in 0.10 M KNO3. • Is the NaCl a source of a common ion or uncommon ions? • Na+ and Cl– are uncommon ions. • This NaCl should increase or decrease the solubility of KOOC(CHOH)2COOH? • Increase. • Is the KNO3 a source of a common ion or uncommon ions? • K+ is a common ion. NO3– is an uncommon ion. • This KNO3 should increase or decrease the solubility of KOOC(CHOH)2COOH? • Decrease. The common ion effect is usually greater than the uncommon ion effect.

  11. CALCULATING Ksp FROM SOLUBILITY • In today’s experiment you will measure the concentration of –OOC(CHOH)2COOH(aq) by titration with standardized sodium hydroxide (NaOH) to a phenolphthalein endpoint. • Potassium hydrogen tartrate is a monoprotic acid; that is, only 1 hydrogen will be neutralized by titration with NaOH. • –OOC(CHOH)2COOH(aq) + OH–(aq) → • –OOC(CHOH)2COO–(aq) + H2O(l)

  12. TITRATION USING PHENOLPHTHALEIN AS AN INDICATOR • Stop adding base when the indicator just begins to turn a faint but stable pink. This is the endpoint.

  13. SAFETY • Give at least 1 safety concern for the following procedure. • Using KOOC(CHOH)2COOH, NaCl, KNO3, NaOH, and phenolphthalein. • These are irritants. Wear your goggles at all times. Immediately clean all spills. If you do get either of these in your eye, immediately flush with water. • Your laboratory manual has an extensive list of safety procedures. Read and understand this section. • Ask your instructor if you ever have any questions about safety.

  14. SOURCES • McMurry, J., R.C. Fay. 2004. Chemistry, 4th ed. Upper Saddle River, NJ: Prentice Hall. • Petrucci, R.H. 1985. General Chemistry Principles and Modern Applications, 4th ed. New York, NY: Macmillan Publishing Company. • Traverso M. 2006. Titration using Phenolphthalein as an Indicator. Available: www.chemistry.wustl.edu/.../AcidBase/phph.htm [accessed 14 September 2006].

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